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The Relationship between the Classical
Time Average of a Particle's Position and the
Quantum Mechanical Probability Density Function

There has probably been a sneaking suspicion among many physicists that the probability density functions of quantum mechanics are merely the time average of a body's position rather than reflecting the indetermination of its location. For example, a turning fan or propeller appears to be a blurred disk even though the fan or propeller has a definite position and extent at any instant of time.

When quantum mechanics started getting solutions in the 1920's to physical problems there was a dispute on how to interpret the results. The Copenhagen interpretation was that the solutions pertained to the probability density distributions of particles. This meant that particles no longer had substance per se but instead there were simply probability distributions. Albert Einstein, for one, could not accept this view of reality. He expressed his disagreement by expressing it roughly in the form of "The Old One does not play dice with the world."

The Creation of a Probability Density Function
for a Particle with a Deterministic
Periodic Trajectory

Suppose a particle follows a periodic trajectory given by X(t), where X is a vector of the particle's coordinates. For simplicity let the trajectory be one dimensional.

The probability to be considered is the probability that the particle is in the interval [x-½Δx, x+½Δx] at any randomly selected point in time. The time spent in this interval of length Δx is

Δt = Δx/|v(x)|

where v is the velocity dx/dt at point x. If T is the period of the particle's oscillation then the probability is Δt/T.

As the interval length Δx goes to zero the probability goes to the probability density. Thus the probability density P(x) is 1/(T|v(x)|). The period T is given by

T = ∫(1/|v(x)|)dx

where the integration is over the path of the particle.

Illustration with the
Harmonic Oscillator

A harmonic oscillator is a point particle with mass m subject to a restoring force of −kx. The potential energy at x is ½kx². The total energy E is then

E = ½mv² + ½kx²

The total energy is constant so the velocity v is given by

v(x) = [(E−½kx²)/(½m)]½
which reduces to
v(x) = [(2E/m) − (k/m)x²]½
the term (k/m)
can be factored out
of the expression
to give
v(x) = (k/m)½[(2E/k) − x²]½

The term (k/m)½ is equal to the oscillatory frequency ω.

The extremes of the oscillation are the points where all of the energy is potential and none kinetic. If the extremes are ±xm then

½kxm² = E
and thus
xm = ±(2E/k)½

The previous expression for v can be put into the form

v(x) = (k/m)½[xm² − x²]½
or, equivalently
v(x) = ω[xm² − x²]½
and therefore the
probability density
is proportional to
1/v(x) = 1/ω[xm² − x²]½

Here is a graph showing the shape of the probability density function for a harmonic oscillator for which the extremes are ±1.

There are singularities at ±1.

In order to get the precise expression for the probability density function it is necessary to evaluate the parameter T.

T = ∫(1/|v(x)|)dx = ∫dx/(ω[xm² − x²])½
= (1/ω)∫dx/(xm[1 − (x/xm)²]½)

Let x/xm be denoted as z, so dz=dx/xm. Then the value of T is given by

T = (1/ω)∫-11dz/(1−z²)½

Now let z=sin(θ) so dz=cos(θ)dθ and (1-z²)½=cos(θ). Thus

T = (1/ω)∫-π/2π/2cos(θ)dθ/cos(θ)
T = (1/ω)∫-π/2π/2dθ = (1/ω)[π/2 − (−π/2)] = π/ω

Therefore the classical probability density function pC(x) for the harmonic oscillator is

pC(x) = (ω/π)(1/(ω(xm²−x²))½
= 1/(π(xm²−x²))½)

In particular pC(0)=1/(πxm). Note that the dimensions are right; a probability density for the one dimensional case is per unit length.

The quantum mechanical density function depends upon the energy of the oscillator. Here is an example of an oscillator's probability density function to compare with the classical one. The derivation of the quantum mechanical density function will be dealt with later.

Clearly there is some sort of relationship. It appears that a spatial average of the quantum mechanical probability density function over a suitable small interval would coincide with the classical probability density function. However visual impression is not enough. The proposition appears correct in the middle of the interval but is less certain near the end points. What is needed is the quantitative relations even to confirm the proposition for this illustration. Ultimately it is a mathematically rigorous proof that is desired.

The Derivation of the Quantum
Mechanical Probability Density
Functions for the Harmonic Oscillator

The Hamiltonian function for the harmonic oscillator is just the total energy expressed as a function of momentum p and displacement x; i.e.,

H = ½p²/m + ½kx²

When this is converted into the Hamiltonian operator and the result substituted into the time independent Schrödinger equation one obtains

−½(h²/m) (∂²φ/∂x²) + ½kx²φ = Eφ

where E is the total energy, φ is the wave function for the system and h is Planck's constant divided by 2π.

In solving this equation it is found that the energy is quantized and is given by

En = hω(n+½)

where n is an integer which is called the quantum number for the system and ω is the natural frequency of the system is equal to (k/m)½.

The solutions for the wave function φ are given most conveniently in terms of a dimensionless variable ζ= x/σ, where σ is a natural unit of length given by

σ² = h/(km)½
and hence
σ = h½/(km)¼

The solutions to the Schrödinger equation for the harmonic oscillator φn are then of the form

φn(x) = AnHn(ζ)exp(−ζ²/2)

where An is equal to (2nn!√π)−½. Hn stands for the n-th order Hermite polynomial, which are solutions to the differential equation

Hn"(ζ) − 2ζHn'(ζ) + 2nHn(ζ) = 0

Here are some of the Hermite polynomials.

OrderHermite
Polynomial
01
1
24ζ²−2
38ζ³−12ζ
416ζ4−48ζ2+12

Note that the polynomial order of Hn(ζ) is n. Because Hn is a polynomial the variable has to be dimensionless.

The probability density function is just the square of φ, but with the subtlety that to get the probability for an interval dx the interval has to be expressed in terms of dζ=dx/σ. In other words the probability density function in terms of x is φ(x/σ)²/σ.

The solutions for φ and φ² are given below for n=0,1,2,4 and 5.

In the above display the number of peaks in the probability density function is equal to (n+1); the number of troughs (minima) is equal to n. The parameter n is the principal quantum number and the energy of the system is equal to hω(n+½). Thus for high energy systems the probability density function would appear to be almost continuous. Perhaps a better characterization of it would be like the appearance of a propeller viewed with a stroboscopic light. If the stroboscope is flashing at m times the rate of rotation there would appear to be m copies of the propeller each separated by an angle of 2π/m.

(In this case because of the symmetry of the propeller with respect to the center of rotation the stroboscope reveals 4 copies of the propeller but there are two propellers superimposed at each of the two locations shown.)

The form of the probability density functions in terms of ζ is

φn²(ζ) = (1/(2n+2n!√π)Hn²(ζ)exp(−ζ²)

The peaks and troughs occur where d(φn²)/dζ is equal to zero; i.e.,

2Hn(ζ)H'n(ζ)exp(−ζ²) + Hn²(ζ)(−2ζ)exp(−ζ²) = 0
which reduces to
H'n(ζ) −ζHn(ζ) = 0

The Hermite polynomials satisfy a number of relationships, two of which are

H'n(ζ) = nHn-1(ζ)
H'n(ζ) = ζHn(ζ) − Hn+1(ζ)

Thus the peaks and troughs in the probability density functions occur where

ζHn(ζ) = nHn-1(ζ)
or, alternately
Hn+1(ζ) = 0

Viewing cases shown in the previous display again we see that where the probability density function φn+1² has a zero φn² has a peak (relative maximum). (Note that the case of n=3 is not shown.) The points at the two ends of the range are not really zeroes; they correspond to rapid exponential decline.

The peaks and troughs of the probability density functions φn² are determined by Hn²(ζ). The factor of exp(−ζ²) shifts the points slightly but it is the critical points of Hn²(ζ) that are crucial.

Consider now the critical points of Hn²(ζ). The derivative of this function is

2Hn(ζ)Hn'(ζ)
which will be equal to zero
if Hn(ζ) or Hn'(ζ)
or both, are equal to zero.

Thus the points where φn² is equal to zero are points of relative minima. There can be n of these. The number of points where Hn+1(ζ) is equal to zero can be (n+1). These correspond to the relative maxima.

The general shape of the probability density function comes basically from the contest between ζ2n, the dominant term of Hn²(ζ) and exp(−ζ²) in the function φn²(ζ). For small values of ζ, the lowest order term in Hn(ζ), such as the constant term for n even, wins out but ultimately exp(−ζ²) drives the probability density to zero. The product ζ2nexp(−ζ²) reaches its maxima at ζ=±√n.

Here is an illustration of the phenomenon.

Loosely formulated, the conjecture is that

Pn(z) = ∫w(ζ−z)φn²(ζ)dζ
is at least approximately equal to
(1/v(z))/T

where w(x) is a weight function such that ∫w(x)dx is equal to 1. However for large values of n (which corresponds to large values of energy) φn² oscillates so rapidly between relative maxima and minima that the form of the weighting function doesn't matter much. Thus an average of adjacent maximum and minimum will do fine. Since above it was determined that the minima all have a value of zero the average of an adjacent maximum and minimum is just equal to one half of the maximum.

It would be ideal to have a means of computing the relative maxima of φn²(ζ) as a function of n and ζ but that is not available at the present time. However there is the easy case of φn²(0). The comparison of this test case with the classical value could disprove the conjecture or indicate how it should be modified.

The Probability Density Function
at Zero Displacement

It has already been found that the classical probability density function at zero displacement is

pC(0)=1/(πxm)
where xm=(2E/k)½

The quantum mechanical analysis finds that

φn²(0) = 0 for n odd
and
φn²(0) = Hn²(0)/(2nn!√π)
for n even

where
Hn²(0) = [n!/(½n)!]²

By Stirling's approximation of n!=(2πn)½(n/e)n, for n even

φn²(0) = (1/π)(2/n)½

The appropriate quantum mechanical probability density function for comparison with the classical value is PQ(0)=½φn²(0)/σ where

σ² = h/(mω)
or, equivalently
σ² = hω/k.

PQ(0) may be put into the form

PQ(0) = (1/π)(1/(2nσ²))½
and hence
PQ(0) = (1/π)(k/(2nhω))½

The value of the principal quantum number n is related to the energy of the oscillator by the formula

E = hω(n+½) = hωn[(n+½)/n]

Thus

PQ(0) = (1/π)[k/(2E(1+1/(2n)))]½

On the other hand the classical probability density function for the harmonic oscillator at zero displacement is given by

pC(0)=1/(πxm)
where xm=(2E/k)½
and hence
pC(0) = (1/π)(k/2E)½

The ratio of the two probability density functions is then

pC(0)/PQ(0) = (1+1/(2n))½

Thus as n increases the ratio goes to unity.

Conclusion

The analysis of a harmonic oscillator indicates that a spatial average of the quantum mechanical probability density at zero displacement is at least asymptotically equal to the classical time average of the particle's location. The test case result does rule out exact equality because Stirling's approximation of factorials was used and an average of adjacent maxima and minina was used instead of a weighted spatial average. Furthermore that average would apply not at zero but halfway between the points where the maxima and minima occur. The results indicate the problem warrants further investigation. Murray Gell-Mann remarked that

Bohr brainwashed a whole generation of physicists into believing the problem had been solved.

There might be an intermediate ground. Instead of the wave functions indicating only the intrinsic indeterminacy of the particles or only the time average of the particles' locations, the wave functions might combine both. But it is difficult to see how that could be.

(To be continued.)


References:

Richard L. Liboff, Introductory Quantum Mechanics, Holden-Day Inc., San Francisco, 1980.

Barry Spain and M.G. Smith, Functions of Mathematical Physics, Van Nostrand Reinhold Co., London, 1970.


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